# 2017 AIME II Problems/Problem 13

## Problem

For each integer $n\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$.

## Solution

Considering $n \pmod{6}$, we have the following formulas:

Even and a multiple of 3: $\frac{n(n-4)}{2} + \frac{n}{3}$

Even and not a multiple of 3: $\frac{n(n-2)}{2}$

Odd and a multiple of 3: $\frac{n(n-3)}{2} + \frac{n}{3}$

Odd and not a multiple of 3: $\frac{n(n-1)}{2}$

To derive these formulas, we note the following: Any isosceles triangle formed by the vertices of our regular $n$-sided polygon $P$ has its sides from the set of edges and diagonals of $P$. Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of $n$, unless $n$ is even when one set of diagonals (those which bisect the polygon) comes in a group of $\frac{n}{2}$. Three properties hold true of $f(n)$:

When $n$ is odd there are $\frac{n(n-1)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-1}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, and we must divide by 2 for over-count).*

• Another explanation: For any diagonal or side of the polygon chosen as the base of the isosceles triangle, there is exactly 1 isosceles triangle that can be formed. So, the total number of satisfactory subsets is $\dbinom{n}{2}=\dfrac{n(n-1)}{2}.$

When $n$ is even there are $\frac{n(n-2)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-2}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).

When $n$ is a multiple of three we additionally over-count equilateral triangles, of which there are $\frac{n}{3}$. As we count them three times, we are two times over, so we subtract $\frac{2n}{3}$.

Considering the six possibilities $n \equiv 0,1,2,3,4,5 \pmod{6}$ and solving, we find that the only valid solutions are $n = 36, 52, 157$, from which the answer is $36 + 52 + 157 = \boxed{245}$.

## Solution 2 (elaborates on the possible cases)

In the case that $n\equiv 0\pmod 3$, there are $\frac{n}{3}$ equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.

Select a vertex $P$ of a regular $n$-gon. We will count the number of isosceles triangles with their vertex at $P$. (In other words, we are counting the number of isosceles triangles $\triangle APB$ with $A, B, P$ among the vertices of the $n$-gon, and $AP=BP$.)

If the side $AP$ spans $k$ sides of the $n$-gon (where $k<\frac{n}{2}$), the side $BP$ must span $k$ sides of the $n$-gon, and, thus, the side $AB$ must span $n-2k$ sides of the $n$-gon. As $\triangle ABP$ has three distinct vertices, the side $AB$ must span at least one side, so $n-2k \ge 1$. Combining this inequality with the fact that $1\le k<\frac{n}{2}$ and $k\not = \frac{n}{3}$ (as $\triangle ABP$ cannot be equilateral), we find that there are $\lceil\frac{n}{2}\rceil-2$ possible $k$.

As each of the $n$ vertices can be the vertex of a given triangle $\triangle ABP$, there are $\left(\lceil \frac{n}{2} \rceil -2 \right)\cdot n$ non-equilateral isosceles triangles.

Adding in the $\frac{n}{3}$ equilateral triangles, we find that for $n\equiv 0\pmod 3$: $f(n) = \frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n$.

On the other hand, if $n\equiv 1, 2\pmod 3$, there are no equilateral triangles, and we may follow the logic of the paragraph above to find that $f(n)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n$.

We may now rewrite the given equation, based on the remainder $n$ leaves when divided by 3.

Case 1: $n\equiv 0\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n+78$.

In this case, $n$ is of the form $6k$ or $6k+3$, for some integer $k$.

Subcase 1: $n=6k$ Plugging into the equation above yields $k=6\rightarrow n=36$.

Subcase 2: $n=6k+3$ Plugging into the equation above yields $7k=75$, which has no integer solutions.

Case 2: $n\equiv 1\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$.

In this case, $n$ is of the form $6k+1$ or $6k+4$, for some integer $k$.

Subcase 1: $n=6k+1$ In this case, the equation above yields $k=8\rightarrow n=52$.

Subcase 2: $n=6k+4$ In this case, the equation above yields $k=26\rightarrow n=157$.

Case 3: $n\equiv 2\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\frac{n+1}{3}+\left(\lceil \frac{n+1}{2} \rceil -2\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$.

In this case, $n$ is of the form $6k+2$ or $6k+5$, for some integer $k$.

Subcase 1: $n=6k+2$ The equation above reduces to $5k=77$, which has no integer solutions.

Subcase 2: $n=6k+5$ The equation above reduces to $k=-80$, which does not yield a positive integer solution for $n$.

In summary, the possible $n$ are $36, 52, 157$, which add to $\boxed{245}$.